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Thread: Showing existence of an operator

  1. #1
    Senior Member Dinkydoe's Avatar
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    Showing existence of an operator

    I've been given the following question:

    Let $\displaystyle (\mathcal{H},(,))$ be a real Hilbert space with an orthonormal basis $\displaystyle \left\{e_n\right\}_{n=1}^{\infty} $. Let $\displaystyle F:[0,1]\to \mathcal{H}$ be continuous. Show that there exists a positive self-adjoint operator $\displaystyle T\in B(\mathcal{H})$ such that $\displaystyle (Tx,y) = \int_0^1(F(t),x)(F(t),y)dt$ for all $\displaystyle x,y\in \mathcal{H}$.

    Show moreover that $\displaystyle T$ is compact.

    (you may use that $\displaystyle \int_0^1\sum_{n=N+1}^{\infty}|(F(t),e_n)|^2dt\to 0$ as $\displaystyle N\to\infty$)

    I don't quite understand what must be shown here exactly. Must I find such a operator T explicitly and then show it has the desired property. If so, how? I don't quite see where the hint is useful, I guess it's meant for showing compactness of T.

    Any hint or push in the right direction is very much appreciated.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    [This is not a solution, or even a hint, but it might give you a push in the right direction.]

    To get an idea of what is going on here, start by looking at the very simplest situation, when the function F is constant, say $\displaystyle F(t) = z\in\mathcal{H}$. Then $\displaystyle (Tx,y) = (z,x)(z,y)$, so $\displaystyle Tx = (z,x)z$. This is a rank-one operator (you may perhaps have seen the notation $\displaystyle T = z\otimes z$ for such an operator).

    Next, take a slightly more complicated situation, where $\displaystyle F(t) = z_j$ for $\displaystyle (j-1)/n\leqslant z_j<j/n\;(1\leqslant j\leqslant n)$. (Admittedly, that function F is not continuous, but the idea is to use it as a step function giving a Riemann approximation to a continuous function.) Then $\displaystyle (Tx,y) = \frac1n\sum_{j=1}^n(z_j,x)(z_j,y)$. So $\displaystyle Tx = \frac1n\sum_{j=1}^n(z_j,x)z_j$, or $\displaystyle T= \frac1n\sum_{j=1}^nz_j\otimes z_j$. This is again a finite-rank operator.

    Now use the Riemann approximation idea. Given a continuous function $\displaystyle F:[0,1]\to\mathcal{H}$, let $\displaystyle z_j = F(j/n)\;(1\leqslant j\leqslant n)$, and try to show that as $\displaystyle n\to\infty$ the operators $\displaystyle \frac1n\sum_{j=1}^nz_j\otimes z_j$ defined as in the previous paragraph converge in norm to T. Then T will be a norm limit of finite-rank operators, hence compact.
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